Energy Transfer and Thermodynamics - Assignment Example

1) Definition of the four laws of thermodynamics. The laws of thermodynamics are the zeroth law of thermodynamics, the first law of thermodynamics, the second law of thermodynamics, and the third law of thermodynamics. The zeroth law of thermodynamics states that “if two systems are both in thermal equilibrium with a third system, they are also in thermal equilibrium with each other” (Grote & Antonsson, 2009, p.225). It means that if two systems come into contact with one another, then heat exchange occurs so that a thermal equilibrium is achieved. The zeroth law highlights thermal equilibrium among systems as a transitive relation. This law inspired the concept of temperature (Moran & Shapiro, 2006, p. 14) and the development of a thermometer as a tool for measuring temperature (Grote & Antonsson, 2009, p.226). If T(A) = T(B) and T(B) = T(C), Then T(A) = T(C), Where T(A), T(B), and T(C) are temperature for system A, B, and C respectively. The first law of thermodynamics describes the principle of energy conservation. It says that energy is neither created nor destroyed, but changes from one state to another (Wu, 2007, p. 2). It means that the sum of energy of a process within an isolated system remains the same (Roy, 2002, p. 45). The law is expressed mathematically as follow, ΔE = Q + W, Where ΔE is change in internal energy, Q is heat flow heat into or out of the system, and W is work done. A positive sign is used before W to show work done on the system, a negative sign is used when work is done by the system. The second law of thermodynamics states that "in all energy exchanges, if no energy enters or leaves the system, the potential energy of the final state will always be less than that of the initial state" (Razeghi, 2006, p. 508). It means that it is not possible to convert all heat energy into work. In other words, every time energy is transformed or transferred, some of this energy, and at some point all the energy, becomes less useful. The law also implies that entropy of a closed system can either increase or remain constant, but not decrease. S > 0, or S = 0, Where S is entropy. The third law of thermodynamics states that “as a system approaches absolute zero, all processes cease and the entropy of the system approaches a minimum value” (Rauchfuss, 2008, p. 239). As T → 0K , S → 0, Where T is temperature, and S is entropy. 2) Definition of entropy, explanation of what happens to the motion of water molecules when ice melts into water, and what happens to the entropy in this situation. Entropy is basically a measure of the disorder, or randomness, in a thermodynamic system. It is also defined as a measure of the amount of energy in a thermodynamic system that cannot perform work. It is expressed mathematically as, S = k ln W, Where S is entropy, k is Boltzmann’s constant (1.381 * 10-23), and ln W is simply the number of ways the system can be arranged (Pickover, 2008, p. 324). The molecules of water in ice form are normally bound strongly to each other so that water remains a solid (ice). However, as the kinetic energy rises (temperature increases) the molecules become excited and move up and about while colliding with each other. In other words, the molecules are set into kinetic motion (Rauchfuss, 2008, p. 239). At some when the kinetic energy (also temperature) is high enough, the molecules are moving so rapidly that they cannot stay held together and they free away so that ice melts into water. As the ice melts, the entropy increases. 3) Predicting whether entropy will be 0 for the following processes: a. Dry ice melts, entropy will be > 0. b. Water freezes, entropy will be < 0. c. Gasoline evaporates, entropy will be > 0. 4) Calculation of the ΔS for the following reaction, using the thermodynamic data provided. ΔS = ∑Sproducts - ∑Sreactants a. 2NO(g)+O2(g) → N2O4(g), 2NO -> 2 * 210.8 = 421.6 O2 -> 205.1 N2O4 -> 304.3 ΔS = 304.3 - (421.6 + 205.1) = - 322.4 J/K ΔS = - 322.4 J/K b. 3H2(g) + N2(g) → “NH3(g), 3H2 -> 3 * 130.7 = 392.1 N2 -> 191.6 NH3 -> 192.5 ΔS = 192.5 – (392.1 + 191.6) = = 192.5 - 583.7 = -391.2K/J ΔS = -391.2K/J c. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). CH4 -> 186.3 2O2 -> 2 * 205.1 = 410.2 CO2 -> 213.7 2H2O(l) -> 2 * 69.9 = 139.8 ΔS = (213.7 + 139.8) - (186.3 + 410.2) ΔS = 353.5 - 596.5 = -243 K/J ΔS = - 243 K/J 5) Testing the understanding of temperature measurements and temperature scales. a. Body temperature is 37°C, it is as follows in Kelvin, Fahrenheit and Rankine scales: Assuming TC, TK, TF, and TR is temperature in degree Celsius (°C), Kelvin, Fahrenheit, and Rankine respectively, TK = TC + 273.15 = 37°C + 273.15 = 310.15K TF = TC * 9/5 + 32, = 37°C * 9/5 + 32 = 98.6°F TR = TF + 459.67 = 98.6°F + 459.67 = 558.27°R b. Absolute zero in Celsius, Fahrenheit and Rankine scales: For Celsius -> −273.15 °C For Fahrenheit -> −459.67 °F For Rankine -> 0 °R c. The temperature of a system rises by 35°C during a heating process. Expressing this rise in temperature in Kelvins (K). The temperature rises by 35K. Each degree on the Kelvin scale is the same size as a degree Celsius (Ahrens, 2008. p. 27). d. The temperature of a system rises by 80°F during a heating process. Expressing this rise in temperature in R, K and °C: In Rankine, -> 80°R In Kelvin, -> 80 / 1.8 = 44.44°C In °Celsius, -> 80 / 1.8 = 44.44°C The size of 1 °C and 1 Kelvin is equivalent to the size of 1.8°F. However, the size of a degree for Fahrenheit is equal to the size of one degree of Rankine. 6) The mass flow rate is 4kg/s, the heat of combustion for C3H8 is 46450kJ/kg. Determine the heat release rate. Heat release rate (HRR) = Heat of combustion (Q) * mass flow rate (MFR) HRR = 46450kJ/kg * 4kg/s = 185,800KJ/S = 185,800KJ/S 7) What is Fourier’s Law? Mathematically express Fourier’s Law defining all the terms used within it. What is thermal conductivity? Compare the values of thermal conductivity of metals, insulating materials and gases. What does Fourier’s law have a minus sign? Fourier’s Law is the law that governs heat conduction (Pickover, 2008, p. 2). It says that the rate of heat flow through a unit surface area of a material “is directly proportional to the temperature gradient” perpendicular to the surface (Pickover, 2008, p. 2). Mathematically, it is expressed as follows: Q = - k A dT/dx. Where Q is heat transfer rate, A is area perpendicular to the direction of heat flow, k is thermal conductivity, and dT/dx is temperature gradient in the heat flow direction. The negative sign in the mathematical expression of Fourier’s law shows that the gradient is negative to the direction of heat flow. Thermal conductivity is basically a material property that describes the ability of the material to transfer (conduct) heat. In other words, it is the amount of heat (Q), transferred in time t through a thickness L, in a direction normal to a surface area A as a result of temperature difference, and under stable state conditions (Pickover, 2008, p. 230). Metals have the highest thermal conductivity when compared to insulating materials and gases, while insulating materials have the lowest thermal conductivity (Tabor, 1991, p. 246). Gases have higher thermal conductivities than insulating materials (Tabor, 1991, p. 246). Insulating materials must have low thermal conductivity to serve their purpose. 8) Explain the Stefen-Boltzman Law. What is emissivity? What are the range of values for the emissivity of a surface? Define the terms “black surface” and “grey surface”. What role does the view factor play in determining the rate of heat transfer? What is a blackbody? Stefen-Boltzman Law states that “the total emissive power of a black body is proportional to the fourth power of its absolute temperature” (Barrett & Curtis, 1999, p. 27). In other words, p = σT4, Where p is emissive power, T is the absolute temperature, σ is the Stefen-Boltzman constant (5.6703 10-8 W/m2K4) (Barrett & Curtis, 1999, p. 27). Emissivity is the relative ability of a material’s surface to give off energy through radiation. In other words, it is the ratio of radiation given off by a material to the radiation emitted by a black body at similar temperature (Barrett & Curtis, 1999, p. 27). The range of values for the emissivity of a surface is between zero (0) and one (1). A black surface is a surface of a black body. In other words, it is a surface that absorbs all radiant energy striking on it; or rather, it has an absorptivity equivalent to unity (1) (Pickover, 2008, p. 411). A grey surface is an ideal surface of constant emissivity over all temperatures and wavelengths. The view factor gives the ratio of all radiation that leaves surface 1 and strikes surface 2. A black body is an object (or body) that absorbs all radiation striking on it, or rather, an object with an absorptivity of one (1) (Pickover, 2008, p. 411). 9) Define heat of combustion, heat release rate and combustion reaction giving appropriate equations. Explain the different types of combustion and definitions of the following: Specific heat capacity, latent heat, calorimetry, combustion temperature and chemical equilibrium. Heat of combustion is the heat energy released when a substance goes through complete combustion with oxygen under standard temperature and pressure (Grote, Antonsson, 2009, p. 220). Heat release rate is the heat energy generated per unit time. A combustion reaction is an exothermic reaction involving a compound and oxygen resulting in carbon dioxide and water (Kolesnikov, 2001). Example equations of combustion reaction: CH4 + 2O2 --> CO2 + 2 H2O C10H8 + 12O2 ---> 10CO2 + 4H2O Types of combustion include complete and incomplete combustion. Complete combustion is one where the compound is completely burnt yielding water and carbon dioxide, while incomplete combustion (also known as pyrolysis) occurs where oxygen is inadequate such that the substance is not fully burnt (Kolesnikov, 2001). Specific heat capacity is the amount of heat required to increase the temperature of a unit of mass of a substance by a degree of temperature (Pickover, 2008, p. 213). Latent heat is the heat energy absorbed or released by a substance when it is changing its state, usually without temperature change. This situation is witnessed, for instance, during boiling of water or melting of ice (Wu, 2007). Calorimetry is the measurement of the amount of heat absorbed or released during a chemical process, solution formation, or change of state (Razeghi, 2006). Combustion temperature is simply the lowest temperature required for a combustion reaction to occur (Razeghi, 2006). Chemical equilibrium is a state where there is no net change in the concentrations of the products and reactants over time in a chemical reaction (Razeghi, 2006; Wu, 2007). 10) Determine the rate of heat transfer per unit area for a blackbody at 20°C. Is a good absorber of radiation a good emitter or a poor emitter? (2 Mark). Rate of Heat Transfer (q) = σT4A, where T is the absolute temperature, A is area of the body, σ is the Stefen-Boltzman constant (5.6703 * 10-8 W/m2K4). q = 5.6703 *10-8 W/m2K4 * (20°C + 273.15) K * 1. = 5.6703 * 10-8 * 293.15 * 1 = 1.6622 * 10-3 W A good absorber of radiation is also a good emitter. 11) Does heat depend on the mass of a substance? Does temperature depend on the amount of a substance? (2 Marks). Yes. Heat depends on the mass of a substance. However, temperature does not depend on the amount of a substance; it is an intensive property (Liben, 1987, p. 137). 12) How is natural convection different from forced convection? (2 Marks) In natural convection, the movement of molecules is caused by temperature gradient which affects the density and consequently the buoyancy of a fluid, while in forced convection the movement of molecules occurs as a result of external forces (Ahrens, 2008). 13) Aluminium has a specific heat of 0.902 J/goC. How much heat is lost when a piece of aluminium with a mass of 23.984 g cools from a temperature of 415.0oC to a temperature of 22.0oC? q = m * Cg * (Tf - Ti), Where q amount of heat energy lost, mass of the substance, Cg is specific heat, and Tf - Ti is change in temperature. q = 23.984 * 0.902 J/goC * 22.0oC - 415.0oC = 23.984 * 0.902 * 393 = - 8.5020 * 103 Joules 14) A heat engine draws heat from a combustion chamber at 300°C and exhausts to atmosphere at 10°C. What is the maximum thermal efficiency that could be achieved? Efficiency (η) = 1 – (T2 / T1), Where T1 is supply temperature and T2 is the exhaust temperature. η = 1 – (10 / 300) = 0.9667 The maximum thermal efficiency is 96.67% 15) The temperature of a sample of water increases by 69.5oC when 24 500 J are applied. The specific heat of liquid water is 4.18 J/goC. What is the mass of the sample of water? q = m * Cg * (Tf - Ti), 24,500 = m * 4.18 J/goC * 69.5oC Mass (m) = 24,500 / (4.18 J/goC * 69.5oC) = 24,500 / 290.51 = 84.3344g 16) How much energy does it take to raise the temperature of 70 g of copper by 30 °C? Specific heat of copper is 0.385 J/g ºC. (2 Marks). Energy (q) = m * Cg * (Tf - Ti), = 70g * 0.385 J/g ºC * 30 °C = 808.5 Joules 17) Define the following terms: a) Heat capacity is the amount of heat energy required to change the temperature of a substance by one degree (Tabor, 1991, p.229). b) Specific heat is the amount of heat energy required to increase the temperature of one gram of a substance by one degree Celsius, (or 1 Kelvin) (Pickover, 2008, p. 213). 18) Heat is added to a system, and the system does 26 J of work. If the internal energy increases by 7J, how much heat was added to the system? (2 Marks) Ef – Ei = Q – W, Where Ef – Ei is change in internal energy, Q is heat transferred and W is work done. Therefore, 7J = Q – 26J Q = 7J + 26J = 33J 33 Joules were added to the system. 19) A 60kg block of iron is heated from 22°C to 152°C. How much heat had to be transferred to the iron? (2 Marks) Heat transfer (q) = m * Cg * (Tf - Ti), = (60kg * 1000)g * 0.45J/g 0C * (152°C - 22°C) = 60, 000g * 0.45J/g 0C * 130°C = 3510000 Joules = 3.510 * 103 KJ 3.510 * 103 kilojoules had to be transferred 20) 180J of heat are injected into a heat engine, causing it to do work. The engine then exhausts 40J of heat into a cool reservoir. What is the efficiency of the engine? (2 Marks) Efficiency (η) = heat energy converted to work / heat energy supplied = (180J - 40J) / 180J = 140J / 180J = 0.78 = 78% Efficiency of the engine is 78% 21) What is kinetic energy and how does it relate to the temperature of a system? (2 Marks) Kinetic energy is simply energy of motion (Shipman, Wilson, Todd, 2007, p. 79). Kinetic energy of a body is the energy has because of its motion. It can also be defined as the amount of work required to accelerate an object of a specific mass from rest to a specified velocity (Shipman, Wilson, Todd, 2007, p. 79). Kinetic energy relates to temperature in the sense that temperature of a system indicates the average kinetic energy of the system. In other words, “temperature is a measure of the average speed of the atoms and molecules” (Ahrens, 2007, p.28) in the system. Therefore, the average kinetic energy is directly proportional to temperature. 22) 61.6 ml of milk at 18.6 °C are added to 455.5 ml of coffee at 90.2 °C. What is the final temperature in degrees Celsius of this liquid mixture when thermal equilibrium is reached? Assume coffee has the same properties as pure water. The average density of milk is 1032 kg/m3. The specific heat of milk is 1.97 J/g °C. (2 Marks) At equilibrium qm = qc, Where qm and qC are heat transferred in milk and coffee respectively. Heat transferred (q) = m * Cg * (Tf - Ti), and Mass (m) = density (d) * volume (v). For milk, Density = 1032 kg/m3 = 1.032g/cm3 m = 1.032g/cm3 * 61.6 ml (cm3) = 63.5712g qm = 63.5712g * 1.97 J/g °C * (Tf – 18.6°C) For coffee, q = (Tf – 18.6°C), Density = 1000 kg/m3 = 1g/cm3 m = 1.0g/cm3 * 455.5 ml (cm3) = 455.5g qC = 455.5g * 4.1818 j/g°C * (Tf – 90.2 °C) At equilibrium, qm = qc, 63.5712g * 1.97 J/g °C * (Tf – 18.6°C) = -[455.5g * 4.1818 J/g°C * (Tf – 90.2 °C)] 125.2353 * (Tf – 18.6°C) = - [1904.8099 * (Tf – 90.2 °C)] 125.2353Tf – 2329.3766°C = 1904.8099Tf + 171813.8530 °C) 125.2353 Tf + 1904.8099Tf = 2329.3766 + 171813.8530 2030.0452 Tf = 174143.2296 Tf = 85.7829°C Therefore, the final temperature of the mixture at thermal equilibrium is 85.7829°C 23) 1.5kg of cold water at 4°C is added to a container of 1.5kg of hot water at 65°C. What is the final temperature of the water when it arrives at thermal equilibrium? (2 Marks) q = m * Cg * (Tf - Ti), At equilibrium, qcw = qhw, Where qcw and qhw are heat transfers in cold water and hot water respectively. (1.5kg * 1000)g * 4.1818 j/g°C * (Tf - 4°C) = - [(1.5kg * 1000)g * 4.1818 j/g°C * (Tf - 65°C)] 1500g * 4.1818 j/g°C * (Tf - 4°C) = - [1500g * 4.1818 j/g°C * (Tf - 65°C)] Tf - 4°C = -Tf + 65°C) Tf + Tf = 65°C + 4°C 2Tf = 69°C Tf = 34.5°C Therefore, the final temperature of the water at thermal equilibrium is 34.5°C 24) Gold has a specific heat of 0.129 J/g °C. If 5.00 g of gold absorbs 1.33 J of heat, what is the change in temperature of the gold? (2 Marks) q = m * Cg * (Tf - Ti), 1.33J = 5.00 g * 0.129 J/g °C * (Tf - Ti) (Tf - Ti) = 1.33J / 5.00 g * 0.129 J/g °C = 2.0620°C The change in temperature of the gold is 2.0620°C 25) A gas absorbs 2.5J of heat and then performs 1.5J of work. What is the change in internal energy of the gas? (2 Marks) Ef – Ei = Q – W = 2.5J - 1.5J = 1J The change in internal energy of the gas is 1J 26) Explain the ideal gas law, give the mathematical equation and define all the terms used. (2 Marks) The ideal gas law is a law that governs the relationship between volume, temperature, and pressure of an ideal gas. It states that the amount of an ideal gas is determined by its temperature, pressure and volume (Zumdahl & DeCoste, 2010, p.). It is expressed mathematically as, PV = nRT Where P is pressure, V is volume, n is number of gas’ moles, R is the universal gas constant (8.3145 J/mol K), and T is temperature. 27) Explain what intensive and extensive properties are, giving examples of each to support your answer. (1 Mark) Intensive properties are properties that are not dependent on the quantity of a substance (Liben, 1987, p. 137). An example is the boiling point. Water, for instance, will boil at the same temperature irrespective of how many grams are involved. On the other hand, extensive properties are those that dependent on the amount of a substance (Moran & Shapiro, 2006, p. 28). An example of an extensive property is volume of substance. A gram of water will have less volume than 10 grams of water. 28) Explain Newton’s Law of cooling and give the mathematical equation defining all the terms used. (2 Marks). The Newton’s Law of cooling says that the rate of cooling of an object is directly proportional the temperature difference between it and the surroundings (Pickover, 2008, p. 104). It means that if an object is much hotter than its surroundings then its cooling rate is high such that it cools fast. The law is expressed mathematically as follows (Pickover, 2008, p. 104), dT/dt = - K (T - T8), Where dT/dt is the temperature derivative in relation to time t, T and T8 are temperatures of the object and surroundings respectively, and K is proportionality constant. 29) Discus the different types of systems encountered in thermodynamics. (3 Mark) There are mainly three types of thermodynamic systems: a closed system, an open system, and an isolated system (Shipman, Wilson, Todd, 2007, p. 75). A closed thermodynamic system is one which there no transfer of mass into, or from the surroundings. However, an open system involves exchange of both energy and mass with the surroundings. For an isolated system, there is no influence by the surroundings. Therefore, both mass and energy remains constant in an isolated system. 30) Discuss the variables that are used to quantify a gas. (2 Marks) The variables that are used to quantify a gas are volume, pressure, temperature, and moles. A mole of a gas is the amount of gas that has similar number of elementary units (electrons, ions, atoms, or molecules) as the atoms of 12 grams of carbon-12 (Pickover, 2008, p. 197). The volume gives the spaces occupied by the gas. The pressure of the gas and temperature of the environment are also used. Reference list Ahrens, C. D. 2007. Meteorology today: an introduction to weather, climate, and the environment. Florence, Kentucky: Cengage Learning. Ahrens, C. D. 2008. Essentials of meteorology: an invitation to the atmosphere. Florence, Kentucky: Cengage Learning. Barrett, E.C. & Curtis, L. F., 1999. Introduction to environmental remote sensing. London: Routledge. Grote, K, & Antonsson, E.K. (2009). Springer Handbook of Mechanical Engineering. New York: Springer. Kolesnikov, I. M. 2001. Thermodynamics of spontaneous and non-spontaneous processes. Hauppauge, New York: Nova Publishers. Moran, M.J., & Shapiro, H.N., 2006. Fundamentals of engineering thermodynamics. Chichester, West Sussex: John Wiley and Sons Ltd. Pickover, C. A. 2008. Archimedes to Hawking: laws of science and the great minds behind them. Oxford: Oxford University Press. Rauchfuss, H. (2008). Chemical evolution and the origin of life. Tokyo: Springer Japan. Razeghi, M. (2006). Fundamentals of solid state engineering. Boston, Massachusetts: Birkhäuser. Roy, B. N. 2002. Fundamentals of classical and statistical thermodynamics. West Sussex: Wiley-Interscience. Shipman, J., Wilson, J. D., Todd, A. 2007. An introduction to physical science. Florence, Kentucky: Cengage Learning. Tabor, D. 1991. Gases, liquids, and solids: and other states of matter. Cambridge: Cambridge University Press. Wu, C. (2007).Thermodynamics and heat powered cycles: a cognitive engineering approach. Hauppauge, New York: Nova Publishers. Zumdahl, S.S., & DeCoste, D.J., 2010. Introductory chemistry. Florence, Kentucky: Cengage Learning. Read More